, y = 2 when

Given and

This is a first order linear differential equation of the form

Here, P = –3 cot x and Q = sin 2x

The integrating factor (I.F) of this differential equation is,

We have

[∵ m log a = log a^m]

∴ I.F = cosec³x [∵ e^{log x} = x]

Hence, the solution of the differential equation is,

Recall

⇒ ycosec³x = 2(–cosec x) + c

⇒ ycosec³x = –2cosec x + c

∴ y = –2sin²x + csin³x

However, when, we have y = 2

⇒ 2 = –2(1)² + c(1)³

⇒ 2 = –2 + c

∴ c = 4

By substituting the value of c in the equation for y, we get

y = –2sin²x + (4)sin³x

∴ y = –2sin²x + 4sin³x

Thus, the solution of the given initial value problem is y = –2sin²x + 4sin³x